anonymous
  • anonymous
A grain silo is shown below. What is the volume of grain that could completely fill this silo rounded to the nearest whole number? Use 22/7 for pi. 19,008 ft3 19,461 ft3 6,336 ft3 453 ft3
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
anonymous
  • anonymous
Sum together half of a sphere for the top part, with a cylinder for the bottom part. That's all there is to it :-) Do you know the formulas for a cylinder and a sphere? h = 168 ft r = 6 ft
anonymous
  • anonymous
no i don't

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anonymous
  • anonymous
\[\large V_{sphere} = \frac{4 \pi r ^3}{3}\] \[\large V_{cylinder} = \pi r^2 h\] \[V_{total} = \frac{V_{sphere}}{2} + V_{cylinder}\]
anonymous
  • anonymous
Can you solve it now? ;-D
anonymous
  • anonymous
no i'm sorry i'm completely lost :-(
anonymous
  • anonymous
r = radius (both for the cylinder and sphere) , h = height of the cylinder Separate the curved, hemispherical top from the cylinder under it. You'll just add volume the two shapes together. What are you confused on specifically?
anonymous
  • anonymous
i'm just naturally confused on everything that has to do with maths you have to talk to me like a toddler and tell mi step by step
anonymous
  • anonymous
Can you identify the radius in the drawing? And the height?
anonymous
  • anonymous
the radius for the entire drawing?
anonymous
  • anonymous
There's only one radius, yes. It works for both the sphere and cylinder, like I said.
anonymous
  • anonymous
is it 6 ft?
anonymous
  • anonymous
Yes... |dw:1342123496209:dw|
anonymous
  • anonymous
And height is "h", which is?
anonymous
  • anonymous
168 ft
anonymous
  • anonymous
2/3*22/7*6*6*6 + 1/3*@22/7*6*6*168 = ............
anonymous
  • anonymous
Ok, now scroll up and put r & h into those two equations for the Volume of a Sphere and Cylinder. Substitute, replacing r with "(6 ft)", and replacing h with "(168 ft)".
anonymous
  • anonymous
ok
anonymous
  • anonymous
|dw:1342123834485:dw| find the volume of given figure
anonymous
  • anonymous
@abhishekjha29 no, it's a cylinder under a hemisphere, not a cone.
anonymous
  • anonymous
ya tats ok.but i have a new figure wid a cone instead of culinder
anonymous
  • anonymous
so i use 3/4 3.14 times 6 to the 3rd power to find the volume of the sphere?
anonymous
  • anonymous
??? @abhishekjha29 the volume of a cone = \(\large\frac{\pi r ^2 h}{3}\) , exactly a third of a cylinder's volume with the same dimensions for height and radius.
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
when i do it it tells me that the volume for the cylinder is 6330.24
anonymous
  • anonymous
@rudyjanay I would leave the 3.1415... \(\pi\) out of it until the end, just keep it along like it's a unit like "feet". You can indeed do that above, but keep in mind it will be an estimate because: \(\pi \approx 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 062...\)
anonymous
  • anonymous
Best to just leave it alone until you're all done with the formulas :-D
anonymous
  • anonymous
Kind of like how I did in the cylinder part above, if you see what I did
anonymous
  • anonymous
When you're all done, multiply that big, non-decimal number and it'll give you your messy decimal answer.
anonymous
  • anonymous
ok i'll try leaving the 3.14 out
anonymous
  • anonymous
:-) I think you've got it down now It's just plug & chug once you got the setup up done. The tricky part was recognizing it's half of a sphere (a hemisphere) plus a cylinder added together, and that the cylinder and the hemisphere have the same radius.
anonymous
  • anonymous
oh i guess my phone was giving me the wrong answer because i don't have a calculator with me
anonymous
  • anonymous
the answer that i got is 19 461
anonymous
  • anonymous
I just noticed I made a typo let me correct myself... \[\large V_{cylinder} = \pi r^2 h\] \[\large V_{cylinder} = \pi (6 \ ft)^2 (168 \ ft)\] \[\large V_{cylinder} = \pi (36 \ ft^2)(168 \ ft)\] \[\large V_{cylinder} = 6048\pi \ ft^3\] And back into the total... \[\large V_{total} = \frac{288\pi \ ft^3}{2} + (6048\pi ft^3)\] \[\large V_{total} = 144\pi \ ft^3 + 6048\pi ft^3\] \[\large V_{total} = 6192\pi \ ft^3 \approx \ 19452.74171... ft^3\]
anonymous
  • anonymous
oh so i was right
anonymous
  • anonymous
So yes, you're good. The answer choice you were given is actually an estimate, and not as exact as what I just did :-)
anonymous
  • anonymous
ohhhh great thank you
anonymous
  • anonymous
6292\(\pi\) ft\(^3\) is the exact answer, and what I would be expect to do at work
anonymous
  • anonymous
Because if a construction firm wants to make an estimate to their liking, they can just pull out their pocket calculator or smartphone and see how many decimal places they care about for the project
anonymous
  • anonymous
Sometimes accuracy is critical and a matter of life & death, other times you don't have to be so particular
anonymous
  • anonymous
thats why i don't get maths
anonymous
  • anonymous
Example of where accuracy is a matter of life & death: ammunition factory. Even the smallest screw up could mean a detonation, a dud bullet in a critical field operation that means some cop or soldier gets killed, contracts being canceled etc. Another good example would be a factory where they make care engines. If they get the radius slightly wrong your engine won't work well or is more prone to breakdowns. Math can seem useless until you have something that makes it mean something :-)
anonymous
  • anonymous
Car engines* lol not "care"
anonymous
  • anonymous
are you a math wiz or something
anonymous
  • anonymous
Engineering student. I'm not mathematician, but I love the application of it to stuff that matters :-3 Fire & explosives for example is a specialty of mine, and game design is my hobby (which also uses math quiet a bit because of physics stuff).
anonymous
  • anonymous
Math itself though, meh... I loose focus if it doesn't mean anything practical.
anonymous
  • anonymous
oh ok i'm gonna check on my other question

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